《电路》（英文版）13-4 Hybrid(混合) parameters

§13-4 Hybrid(混合) parameters I1=y11+y1212-s1 yuv J JI I2=y21V+y2V2→12=少1+(y2 J12v21 J V1=h111+h hu, h12, h2p, h22 are h parameters 2=h2 22 The matrix from of h parameter equations: h1h12 h:h12 2 If yn2=y 2l, we have h,2=-h21

§13-4 Hybrid（混合） parameters 2 12 1 11 1 • • • I = y V + y V 2 11 12 1 11 1 1 • • • = − V y y I y V 2 22 1 21 2 • • • I = y V + y V 2 11 12 21 22 1 11 21 2 ( ) • • • = + − V y y y I y y y I 2 12 1 11 1 • • • V = h I + h V 2 22 1 21 2 • • • I = h I + h V h11 , h12 , h21, h22 are h parameters. The matrix from of h parameter equations:               =         • • • • 2 1 21 22 11 12 2 1 V I h h h h I V           =         • • • • 2 1 2 1 V I h I V         = 21 22 11 12 h h h h h If y12 =y 21 , we have h12= -h21. 2 • 1 I • I − + • V 2 − + • V 1

1=h1I1+h12 12=h21I1+h2 LetV2=0句,s the short-circuit input impedance 0 the short-circuit forward current gain I1V2=0 Let /1=0,、v1 the open-circuit reverse voltage gain I1=0 the open-circuit output admittance

2 = 0 • Let V 1 2 0 1 11 = = • • • I V V h 1 2 0 2 21 = = • • • I V I h the short-circuit input impedance the short-circuit forward current gain 1 = 0 • Let I 1 0 2 1 12 = = • • • I V V h 1 0 2 2 22 = = • • • I V I h the open-circuit reverse voltage gain the open-circuit output admittance 2 • 1 I • I − + • V 2 − + • V 1 2 12 1 11 1 • • • V = h I + h V 2 22 1 21 2 • • • I = h I + h V

Example:lfh1=1200,h12=0.0002,h2=50,h220.00005. Findz andz out· Solution: 10-3-800/1=1200I1+2×10-V 800 500(2=)-V2/500501+50×10v 1∠0°m④ 0.001=2000I1+0.0002V 0=50I1+0.00025V2 丿1=0.592n 172 1=0.5101 40 102n I2=20.44 200011 =1160g 0.0002 22.4kQ 丿s=0 2 20001 50/1+0.00005 0.0002

Example: If h11=1200, h12=0.0002,h21=50, h22=0.00005S. Find Zin and Zout. Solution:     = − = +  − = +  • − • • • • − • • − 2 6 2 2 1 2 4 1 1 3 ( ) / 5000 50 50 10 10 800 1200 2 10 I V I V I I V     = + = + • • • • 1 2 1 2 0 50 0 00025 0 001 2000 0 0002 I . V . I . V or     = − = • • V mV V mV 102 0.592 2 1     = = • • I A I A   20.4 0.510 2 1 172 1 2 • = − • V V 40 1 2 • = • I I = • =  • 1160 1 1 I V Zin =  + − − = = = • • • • • • k I I I V s I V Zout 22.4 ) 0.0002 2000 50 0.00005( 0.0002 2000 0 1 1 1 2 2 5000 − + • V 2 I 2 • I 1 • − + • V 1 − + 800 1 0 mV  